mon to the triangles EGC, FGC, the two sides, EG, GC are equal to the two, FG, GC, each to each; and the base CE is equal (I. def. 30) to the base CF; therefore the angle CGE is equal (I. 8) to the angle CGF; and they are adjacent angles: therefore CG is perpendicular (I. def. 8) to AB. Hence, from the given point Ca perpendicular CG has been drawn to the given line AB: which was to be done.

PROP. XIII. THEOR.-The angles which one straight line makes with another upon the one side of it are together equal to two right angles.

Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD; these are together equal to two right angles. For, if the angle CBA be equal to ABD, each of them (I. def. 8) is a right angle. But if not, from the point B draw (I. 11) BE at right angles to CD:* therefore the angles CBE, EBD, are two right angles. Then (I. ax. 10) DBE is equal to the two angles DBA, ABE together; add

:

C

B

D

E A

C

B

D

EBC to each of these equals therefore the angles CBE, EBD, are equal (I. ax. 2) to the three angles DBA, ABE, EBC. Again (I. ax. 10), the angle CBA is equal to the two angles CBE, EBA; add ABD to these equals: therefore the angles CBA, ABD are equal to the three angles CBE, EBA, ABD. But CBE, EBD have been demonstrated to be equal to the same three angles: therefore (I. ax. 1) the angles CBE, EBD are equal to the angles CBA, ABD; but CBE, EBD are two right angles: therefore CBA, ABD are together equal to two right angles. Wherefore the angles which one straight line, &c.

Cor. 1. From this it is manifest, that if two straight lines cut one another, the four angles which they make at their point of intersection, are together equal to four right angles.

Cor. 2. And consequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

PROP. XIV. THEOR.-If, at a point in a straight line, two other straight lines on the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line.

At the point B in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles

* The substance of the remainder of this demonstration is, that the angles ABC, ABD are together equal to EBC, EBD, each pair being equal to the three angles CBE, EBA, ABD.

ABC, ABD equal together to two right angles: BD is in the same straight line with CB.

C

R

D

E

For if BD be not in the same straight line with CB, let BE be in the same straight line with it. Therefore, because AB makes angles with the straight line CBE, upon one side of it, those angles ABC, ABE, are together equal (I. 13) to two right angles; but the angles ABC, ABD are likewise together equal (hyp.) to two right angles: therefore (I. ax. 11 and 1) the angles CBĂ, ABE are equal to the angles CBA, ABD: take away the common angle ABC, and the remaining angle ABE is equal (I. ax. 3) to the remaining angle ABD, a part to the whole, which is absurd. Therefore BE is not in the same straight line with BC: and in like manner it may be demonstrated, that no other line can be in the same straight line with it but BD, which therefore is in the same straight line with CB. Wherefore, if at a point, &c.

Cor. If, at a point in a straight line, two other straight lines meet on the opposite sides of it, and make equal angles with the parts of it on opposite sides of the point, the two straight lines are in one and the same straight line.

Let AEB (fig. to the next proposition) be a straight line, and let the angles AEC, BED be equal; CE, ED are in the same straight line. For, by adding the angle CEB, to the equal angles AEC, BED, we have BED, BEC together equal to AEC, CEB, that is (L. 13) to two right angles; and therefore, by this proposition, CE, ED are in the same straight line.

PROP. XV. THEOR.-If two straight lines cut one another, the vertical, or opposite angles are equal.

Let the two straight lines AB, CD cut one another in E; the angle AEC is equal to the angle DEB, and CEB to AED.

C

E

Because the straight line AE, makes with CD the angles CEA, AED, these angles are together equal (I. 13) to two right angles. Again, because DE makes with AB the angles AED, DEB, these also are together equal to two right angles; and CEA, AED have been Ademonstrated to be equal to two right angles; wherefore (I. ax. 11 and 1) the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle AED, and (I. ax. 3) the remaining angles CEA, DEB are equal. In the same manner it can be demonstrated that the angles CEB, AED are equal. Therefore, if two straight lines, &c.

It

* In the proof here given, the common angle is AED; and CEB might with equal propriety be made the common angle. In like manner, in proving the equality of CEB and AED, either AEC or BED may be made the common angle. is also evident, that when AEC and BED have been proved to be equal, the equality of AED and BEC might be inferred from the 13th proposition, and the third

axiom.

PROP. XVI. THEOR.-If one side of a triangle be produced, the exterior angle is greater than either of the interior remote angles.

Let ABC be a triangle, and let its side BC be produced to D; the exterior angle ACD is greater than either of the interior remote angles CBA, BAC.

Bisect (I. 10) AC in E, join (I. post. 1) BE, and produce it (I, post. 2) to F; make (I. 3) EF equal to BE; and join FC.

B

A

E

HC

Because (const.) AE is equal to EC, and BE to EF; the two sides, AE, EB are equal to the two, CE, EF, each to each; and the angles AEB, CEF are equal (I. 15), because they are opposite, vertical angles: therefore the angle BAE is equal (I. 4, part 3) to ECF; but the angle ECD is greater (I. ax. 9) than ECF: therefore the angle ACD is greater than BAE.

In the same manner, if the side AC be produced (I. post. 2) to G, and BC be bisected in H, and if a straight line be drawn from A to H, and continued till the produced part be equal to AH, and if the extremity of that part be joined with C, it may be demonstrated that the angle BCG, that is (I. 15) ACD is greater than ABC. Therefore, if one side, &c.

Cor. 1. Hence from a point without a straight line, only one perpendicular can be drawn to the line. For, if there could be two, a triangle would be formed which would have an exterior angle equal to an interior and remote one, each being a right angle; which, by the proposition, is impossible.

Cor. 2. If from a point without a line, two straight lines be drawn cutting it, one of them at right angles to it, and the other not, the perpendicular one will fall on that side of the other on which the acute angle is: for if it fell on the other, the exterior acute angle would be greater than the interior and remote right angle, which (I. def. 10) is impossible.

PROP. XVII. THEOR.*-Any two angles of a triangle are together less than two right angles.

Let ABC be a triangle: any two of its angles are together less than two right angles.

A

Produce BC to D. Then, because ACD is the exterior angle of the triangle ABC, it is greater (I. 16) than the interior and remote angle ABC. To each of these add ACB: therefore the angles ACD, ACB are greater (I. ax. 4) than ABC, ACB: but ACD, ACB are together equal (I. 13) to two right angles: therefore the angles ABC, BCA are less than two right

B

C

D

This proposition might be omitted, as it is included in the 32d of this book, and

is not referred to in any thing preceding that proposition.

angles. In like manner it may be demonstrated, that BAC, ACB are less than two right angles; and if CB be produced through B, the same may be demonstrated respecting the angles A and B. Therefore any two angles, &c.

Cor. Hence every triangle must have at least two acute angles.

PROP. XVIII. THEOR.*—If two sides of a triangle be unequal, the greater side has the greater angle opposite to it.

D

Let ABC be a triangle, of which the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA. Because AC is greater than AB, make (I. 3) AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater (I. 16) than the interior and remote angle C; but (I. 5) ADB is equal to ABD, because the sides AB, AD are equal: therefore the angle ABD is likewise greater than C; wherefore much more is the angle ABC greater than C. Therefore, if two sides, &c.

B

C

PROP. XIX. THEOR.-If two angles of a triangle be unequal, the greater angle has the greater side opposite to it.

Let ABC be a triangle, of which the angle B is greater than C; the side AC is likewise greater than AB.

A

For if it be not greater, AC must either be equal to AB, or less than it. It is not equal, because then (I. 5) the angle B would be equal to the angle C; but it is not. Neither is it less; because then (I. 18) the angle B would be less than the angle C, but it is not. AC, therefore, is not less than AB; and it has been shown that it is not equal to AB: therefore AC is greater than AB. Wherefore, if two angles, &c.

B

PROP. XX. THEOR.-Any two sides of a triangle are together greater than the third side.†

*Let the learner compare this proposition and the following with the 5th and 6th of this book.

This proposition might be proved with equal facility by producing AB through B, till the whole line thus produced is equal to AC, and joining C with the extremity of the produced part. The following proof is very neat and easy. From AC cut off AD equal to AB; bisect (I. 9) the angle BAC by the straight line AE, and join ED. Then, in the triangles BAE, DAE, BA is equal to DA, AE common, and the angle BAE equal to DAE: therefore (I. 4, part 3) the angle ADĚ is equal to B. But (I. 16) ADE is greater than C; therefore also B is greater than C.

B

C

The truth of this proposition is so manifest, that a proof of it is given merely to avoid increasing the number of axioms. If, as is done by Archimedes, a straight line were defined to be the shortest distance between two points, this proposition would follow immediately as a corollary.

Let ABC be a triangle ; any two sides of it are together greater than the third side.

Produce (I. post. 2) BA to the point D, making (I. 3) AD equal to AC; and join DC.

Because DA is equal to AC, the angle ADC is likewise equal (I. 5) to ACD; but the angle BCD is greater (I. ax. 9) than ACD: therefore the angle BCD is greater than ADC: and because in the triangle DBC, the angle BCD is greater than BDC, and since (I. 19) the greater side is opposite to the greater angle: therefore the side DB is greater than the side BC; but DB is equal to BA and AC; therefore BA, AC are greater than BC. In the same manner, it might be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB. Therefore, any two sides, &c.

B

C

Cor. Hence the difference of any two sides of a triangle, is less than the third side. For, since BA and AC are together greater than BC, from each of these take AC; then (I. ax. 5) BA is greater than the excess of BC above AC.

PROP. XXI. THEOR.*—If from a point within a triangle, two straight lines be drawn to the extremities of one of the sides, these lines are together less than the other sides, but contain a greater angle.

From the point D, within the triangle BAC, let the straight lines DB, DC be drawn to the extremities of BC; DB, DC are less than AB, AC, but the angle BDC is greater than BAC.

DE

Produce (I. post. 2) BD to E. Then, because two sides of a triangle are greater (I. 20) than the third, the two sides BA, AE are greater than_BE. To each of these add EC: therefore (I. ax. 4) BA, AC are greater than BE, EC. Again, because (I. 20) the two sides CE, ED are greater than CD, add DB to each: therefore CE, EB are greater than CD, DB. But it has been shown that BA, AC are greater than BE, EC: much more then are BA, AC greater than BD, DC.

B

Again, the exterior angle BDC of the triangle CDE is greater (I. 16) than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than CEB: much more then is the angle BDC greater than BAC. Therefore, if from a point, &c.

A proof as easy as the one given above would be obtained by bisecting the angle BAC; as it follows from the 16th proposition, that each of the angles made by the bisecting line with BC is greater than the half of BAC; and therefore (I. 19) each segment of BC is less than the adjacent side.

This proposition is never referred to by Euclid, except in the 8th proposition of the third book; and that proposition may be proved without it.

By means of the 32d of this book, it may be proved that the angle BDC exceeds BAC by the sum of the angles ABD, ACD.